NTA Abhyas JEE Main2020PhysicsAtomic PhysicsPractice
For sodium light, the two yellow lines occur at λ 1 and λ 2 wavelengths. If the mean of these two is 6000 A ∘ and λ 2 - λ 1 = 6 A ∘ , then the approximate energy difference between the two levels corresponding to λ 1 and λ 2 is
Options
- A2 × 10 - 3 e V
- B2 e V
- C2000 e V
- D2 × 10 - 6 e V
Correct answer
A. 2 × 10 - 3 e V
Step-by-step solution
Given, mean of λ 1 and λ 2 = 6000 Å λ 2 - λ 1 = 6 Å i.e. λ 1 + λ 2 2 = 6000 Å λ 1 + λ 2 = 12000 Å ...(i) λ 2 - λ 1 = 6 Å ...(ii) Equating Equation (i) and (ii), we get 2 λ 1 = 12006 λ 2 = 12006 2 = 6003 Å and λ 1 = 5997 Å Now, the energy difference is ∆ E = h c λ 1 - h c λ 2 ∆ E = h c 1 λ 1 - 1 λ 2 ∆ E = 6.6 × 10 - 34 × 3 × 10 8 × λ 2 - λ 1 λ 1 λ 2 = 6.6 × 10 - 34 × 3 × 10 8 × 6 × 10 - 10 5997 × 10 - 10 × 6003 × 10 - 10 × 1.6 × 10 - 19 e V = 2.12 × 10 - 3 e V ≃ 2 × 10 - 3 e V